How do I chase the identity around this commutative map to show this equality?

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From Mac Lane's Category Theory:

I can see how the chasing $1_{Fx}$ around the right of graph $(3)$ goes $1_{Fx} \rightarrow \eta_x \rightarrow L \circ \eta_x$, but how is $\eta'_x \circ L$ found by chasing around the left side of $(3)$?

It seems like it should just be $1_{Fx} \rightarrow K(1_{Fx})=1_{KFx} \rightarrow \eta'_x$.

What am I misunderstanding here?


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First, $L\circ\eta_x$ doesn't make sense, you would get $L(\eta_x)$ usually written without parenthesis as $L\eta_x$. Similarly, $\eta_x\circ L$ doesn't make sense. $L$ is not an arrow of $X$. The notation $L\mathscr{M}$ is ambiguous depending on what $\mathscr{M}$ is. If $\mathscr{M}$ is a functor, it means $L\circ\mathscr{M}$. If it is an object or arrow in the domain of $L$, it means $L(\mathscr{M})$. If it is a natural transformation, it means whiskering which corresponds to the horizontal composition $id_L*\mathscr{M}$. Via typographical conventions and context, this should always be clear. (Note that this means $L\eta_x$ could mean either $(L\eta)_x$ or $L(\eta_x)$, but these are the same thing by the definition of horizontal composition.)

You are right that you get $1_{KFx}=1_{F'Lx}$, but the image of this when $\varphi'_{Lx,KFx}=\varphi'_{Lx,F'Lx}$ is applied is not $\eta'_x$, it is $\eta'_{Lx}$. $\eta'_x : x\to G'F'x$ which is not an element of $X'(Lx,G'KFx)=X'(Lx,G'F'Lx)$. $\eta'_{Lx}=(\eta'L)_x$ by definition of horizontal composition/whiskering.